1) √ (15-x) + √ (3-x) = 62) √ (x+8) >x+2

1) √ (15 - x) + √ (3 - x) = 6;

a) ОДЗ:

{15 - x ≥ 0;

{3 - x ≥ 0; {x ≤ 15;

{x ≤ 3; x ∈ (-∞; 3].

b) Возводим в квадрат:

15 - x + 2 * √ (15 - x) * √ (3 - x) + 3 - x = 6^2; 2 * √ (15 - x) * √ (3 - x) = 2x + 18; √ ((15 - x) (3 - x)) = x + 9; (15 - x) (3 - x) = (x + 9) ^2; 45 - 18x + x^2 = x^2 + 18x + 81; 36x = - 36; x = - 1 ∈ (-∞; 3].

2) √ (x + 8) > x + 2;

a)

{x + 2 < 0;

{x + 8 ≥ 0; {x < - 2;

{x ≥ - 8; x ∈ [-8; - 2).

b)

{x + 2 ≥ 0;

{x + 8 > (x + 2) ^2; {x ≥ - 2;

{x + 8 > x^2 + 4x + 4; {x ≥ - 2;

{x^2 + 3x - 4 < 0.

Корни трехчлена: x1 = - 4; x2 = 1.

{x ∈ [-2; ∞);

{x ∈ (-4; 1); x ∈ [-2; 1).

Ответ:

1) - 1; 2) [-8; 1).